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If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it height in feet after t second is given by y=45t−16t2. Find the average velocity for the time period begining when t=2 and lasting. 0.1 seconds?

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Respuesta :

HomertheGenius
y = 45 t - 16 t²
f ( 2 s ) = 45 · 2 - 16 · 4 = 90 - 64 = 26 ft
f ( 2.1 s ) = 45 · 2.1 - 16 · 4.41 = 94.5 - 70.56 = 23.94 ft
Average velocity: 
v = ( 23.94 - 26 ) / ( 2.1 - 2 ) = - 2.06 ft / 0.1 s = - 20.6 ft/s

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